Here, we have a common factor 3 in all the terms.

∴    9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy +9]
  We find that      3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy+9)
i.e. a common factor in both the groups does not exist,
 Thus, 3x + 6y + 2xy + 9 cannot be factorised.
On regrouping the terms, we have
   3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y
                            = 3(x + 3) + 2y(x + 3)
                            = (x + 3) (3 + 2y)
Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]
Thus, 9x + 18y + 6xy + 27 = 3(x+3) (2y+3)

We have    12x =
                   36 =

∴        12x + 36 = [(223)x] + (223) 3
                         =

We have                   22y = 2 × 11 × y
                                33z = 3 × 11 × z

∴                     22y - 33z = [2 × (11) × y] + [3 × (11) × z]
                                      
                                      = (11)[2 × y - 3 × z]
                                      = 11[2y - 3z]

We have            14pq = 2 × 7 × p × q = 2 × (7 × p × q)
                         35pqr = 5 × 7 × p × q × r = 5 × (7 × p × q) × r

∴          14pq + 35 pqr = 2 × (7 × p × q) + 5 × (7 × p × q) × r
                                   = (7 × p × q)[2 + 5 × r]
                                   = 7pq(2+5r)